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Evaluate the integral.

$ \displaystyle \int \frac{x \ln x}{\sqrt{x^2 - 1}}\ dx $

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Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 5

Strategy for Integration

Integration Techniques

Missouri State University

Campbell University

University of Michigan - Ann Arbor

University of Nottingham

Lectures

01:53

In mathematics, integration is one of the two main operations in calculus, with its inverse, differentiation, being the other. Given a function of a real variable, an antiderivative, integral, or integrand is the function's derivative, with respect to the variable of interest. The integrals of a function are the components of its antiderivative. The definite integral of a function from a to b is the area of the region in the xy-plane that lies between the graph of the function and the x-axis, above the x-axis, or below the x-axis. The indefinite integral of a function is an antiderivative of the function, and can be used to find the original function when given the derivative. The definite integral of a function is a single-valued function on a given interval. It can be computed by evaluating the definite integral of a function at every x in the domain of the function, then adding the results together.

27:53

In mathematics, a technique is a method or formula for solving a problem. Techniques are often used in mathematics, physics, economics, and computer science.

04:01

02:04

05:10

Evaluate the integral.…

01:57

03:51

02:02

01:38

Evaluate the integrals.

02:50

00:33

01:09

Find the integral.$\in…

01:52

Evaluate the indefinite in…

02:27

Let's start here by taking a U substitution. Then we could find the U and let's go. We don't have a one over X in the original Instagram. So let's go ahead and solve this equation here for X so we can write. This is one over you d x and then take this equation right here the circled one and then solve for DX and you get you a deal. So now let's rewrite this integral in terms of you. First you see and X here. Well, fuse this. That's just eat ofyou. Then you see natural log of X. That's just you. You see d X. That's either the ugo and then in the bottom. We have square root, we see X Square. They're so square both sides of this to get either the to you minus one, and that much is go ahead and clean this up a little bit. That's either the to you over radical E to the to you and then minus one in the radical look. And now, at this point, let's go ahead into another use up here. This time, let's do the equals either. That's who you are so that Devi over two equals either the to you, do you which we can write as VD you and then solve this equation for do you? And if we saw this for you, What for? Do you What are we getting so Devi over to with either the to you. But that's just V so d'You evils Devi over to the So if I come back over here actually, I don't need this last step here. Let me not right that because if I just write, the reason I don't need to do this last step here is because either the to you to you is already and are intolerant here. So if I just write, let me pull up the one half from the DV over too in a girl and then this fry here. Since I pull off the one half, that's just the TV. However, we see that there's a you are here in the front. So to find you, we solve this equation here and then divide by two. So, Ellen, be there as a B over two equals you. So that's L m G over to all over there in the numerator and then in the bottom while luscious V minus one based on our definition of you. So the next step is to just combined these two's here. Let's write that as a four going on the next page. That's one fourth in a girl. And then we had Ellen d the minus one. So it took us to use institutions to get here and yet again there's another integral However, we are almost there. But this inner girl will require by parts. So here, let me use the letter you again. But this time for this formula, right here for the parts El Andy one over D D V. So that's a you there on the left of the on the right. And then since I'm already using V, let me use W here. This will be the remaining part. This is my DW Oops, d w Here. So let me rewrite. This is V minus one to the negative one half and then use your power rule from Cal here. So add one. So we get one half and then divide by half. So he multiplied by two and then we have one over four and then for the inner rule. I'm sorry. Here I should have used up you not be so from I integrated that so Lou V minus in a girl. And then here. That's a w A cz. Well, so that's our formula for integration. My parts again. It's amusing, W because the problem that we're working on has visa in it. So now we go ahead and we will supply you and be together minus integral. And then after this, we have a to and then V minus one one half over the Devi. So yet again we see another integral here. Before we evaluate this, let's just distribute the one fourth through the apprentices over two. And then we could pull out this, too, and then cancel it with one of these twos from the four. So that's minus one half the mighty ass. One to the one half over the. So here I am running out of room. Let me go on to the next page. And then we had distributed the one four. So we were at this step right over here, one half over the DVD. Okay, so we're still not dear. We've already evaluated, like, three in the rules, but yet again, here's another inaugural. This is much simpler than what we've seen so far So But this will be another use of here. You can take t to be V minus one, but inside the radical then if I do dt, that's one. Over here I have to be minus one. So this is your chain rule and then I could rewrite This is one over to TV So that means Devi is to tt. So if I want to evaluate this integral V minus one to the one half over the now I rewrite this all in terms of tea. So this right here, the minus one to the one half That's just he buy this definition up here and then Devi that's just too t d t. We work that out on the side over here and then we're dividing by the But you get by solving this equation. She's very close one. So here we can do if you notice this numerator is tooth where? So you have polynomial divided by polynomial They have the same degree. So you should do polynomial division here. So before you knew partial fraction the composition. So here we should end up with after pulling up the two one minus one over T squared plus one. And then this is just who t minus two tan inverse of tea. You may have memorized this tan inverse part from the table. If not, no worries. The way to get there. Let's just take this and use the subject tricks up t equals tan data and you will have to work through that in a girl. But after after you back substance who you shouldn't have to draw the triangle for this one in particular. Fast substance Who, entity? And there's your answer. But don't forget that was just the integral here. So we finally evaluated this integral. Let me get some color up here. So this is the integral we're working on. We evaluated it. It's in some variable t. We need to get back to you, V. So let's and then also, don't forget this negative one half that I haven't written it yet. So you're going to multiply this expression down here by the negative one half. So the minus one one half Ellen be over to that. This part stays is him. And then after we multiply the bottom right corner by negative. One half that's minus T, but then plus ten in adversity. And now, since we've evaluated all the inner girls, let's go ahead and add that constant C So here the last few steps, or just to make sure we write everything back in terms of X. So for now, let's just leave that be first. Let's get he in terms of the so the minus one one half Ellen be over too, minus he, which is square root B minus one from over here, plus ten. Inverse radical. Okay, now we have everything in terms of the Let's go on to the next page. So let's rewrite this. We're almost there. One last step. Well, we could jump from the directly back to X and we could bypass you and I'LL show you how so we define the earlier by U sub. But originally the you came from. So let's rewrite the in terms of X so obvious either the to you, that's either the two Alan X. That's e to the L. A X Square. And since the exponential function and log function and Versace, when you compose them, they cancel each other out. So this just gives you back X, and then we have a square out here. So that's how you evaluate express the in terms of X. So the last step here. So just replace all the bees with X where you could even yeah, you know, absolute value at all, because X squared is not never negative, if that's your final answer.

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